How does one derive probability densitys involving fractions?
For example, let $X^2$ and $Y^2$ be exponentially distributed random variables with parameter $\lambda = 1$. Determine the PDF for $Z = Y^2/X^2$.
There's the following formula in my book, although I am not quite sure what to do with the integration limits?
$$\int p(x,zx)x dx$$
Also I don't have any multivariable density function?
On
Assuming $X$ and $Y$ are independent, the distribution of $\dfrac{Y^2}{X^2}$ (for $z>0$) is given as:
$$ \begin{eqnarray*} P\left(\dfrac{Y^2}{X^2}<z\right){}={}P\left(Y^2<z\,X^2\right)&{}={}&P\left(Y<\sqrt{z}\,X\right)\,,\,\,\,\,\,\,\mbox{ since }Y\,,\,X\,,z>0\newline &{}={}&\int\limits_{0}^{\infty}P\left(Y<\sqrt{z}\,x\,\bigg|\,X=x\right)f_{X}(x)\,\mathrm{d}x\newline &{}={}&\int\limits_{0}^{\infty}P\bigg(Y<\sqrt{z}\,x\bigg)f_{X}(x)\,\mathrm{d}x\newline &{}={}&\int\limits_{0}^{\infty}\bigg(1-e^{-\sqrt{z}\,x}\bigg)e^{-x}\,\mathrm{d}x\newline &{}={}&\left(1-\dfrac{1}{1+\sqrt{z}}\right){\bf{1}}_{\left\{z>0\right\}}\,. \end{eqnarray*} $$
Therefore, the density of interest is obtained by differentiation:
$$ f_{Z}(z){}={}\dfrac{\mathrm{d}}{\mathrm{d}z}P\left(\dfrac{Y^2}{X^2}<z\right){}={}\dfrac{1}{2}\dfrac{{\bf{1}}_{\left\{z>0\right\}}}{\sqrt{z}\left(1+\sqrt{z}\right)^2}\,. $$
Consider the function $g(x,y)=\left(\dfrac{y}{x},x\right)$. Define $u=\dfrac{y}{x}$ and $v=x$. Then $uv=y$. Then the inverse of $g$ will be $h(u,v)=(v,uv)$. Let's call the pdf of a random variable by $f$ (you call it $p$). By the transformation theorem we know that $f_{ZX}(z,x)=f_{XY}(h(z,x))|\mathbf J(h)|$, where $|\mathbf J(h)|$ is the absolute value of the jacobian determinant of the $h$ transformation. Therefore $f_{ZX}(z,x)=f_{XY}(x,zx)|x|$. Now you want $f_Z$, this is done by calculating it's marginal distribution from $f_{ZX}$, so:
$$f_Z=\int f_{XY}(x,zx)x\,dx.$$
With respect to the multivariate density function I think $X^2$ and $Y^2$ should be independent, this would lead to $f_{XY}=f_Xf_Y$, which you know. If this is the case, the limits of integration of the last integral should be from $0$ to $\infty$, since that is the range for which the exponential distribution is defined.